What is the tangent of f(x)=x^2+10x,\at x=8 ?

The tangent of f(x)=x^2+10x,\at x=8 is y=26x-64

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tangent of f(x)=x^2+10x,\at x=8

tangent of

f (x) = x^{2} + 10 x, at x = 8

y = 26 x - 64

Solution steps

Find the tangent point:

(8, 144)

Find the slope of

f (x) = x^{2} + 10 x : \frac{df ( x )}{d x} = 2 x + 10

Compute the slope of

f (x) = x^{2} + 10 x

x = 8 : m = 26

Find the line with slope m=

26

and passing through

(8, 144) : y = 26 x - 64

y = 26 x - 64

x \to \infty lim (x^{2} e^{2 x})

x \to - 2 lim (\frac{x ^{3} + 1}{x + 1})

y^{^{''}} - 3 y^{^{'}} + 2 y = 14 sin (2 x) + 18 cos (2 x)

\frac{d}{d x} (3 x^{5} - 2 x^{4} - 4 x^{3})

(y - x^{2} + y^{2}) d x - x d y = 0

What is the tangent of f(x)=x^2+10x,\at x=8 ?
The tangent of f(x)=x^2+10x,\at x=8 is y=26x-64